Selection and Adversary Arguments. COMP 215 Lecture 19

Transcription

1 Selection and Adversary Arguments COMP 215 Lecture 19

2 Selection Problems We want to find the k'th largest entry in an unsorted array. Could be the largest, smallest, median, etc. Ideas for an n lg n algorithm? We will think about: Largest, smallest Largest AND smallest Second largest Kth.

3 Finding the Largest Item We have seen a simple algorithm that requires n 1 comparisons in the worst case. That's optimal.

4 Largest and Smallest We could run the previous algorithm twice, which would give us 2n 2 comparisons. We have seen an algorithm that requires comparisons in the worst case. Anyone remember? Is this optimal? First idea is to consider a decision tree. There must be at least n leaves. Therefore the height must be at least lg n. Obviously not a very tight bound. 3 n 2 2

5 Adversary Arguments We design an adversary that forces any algorithm to do as much work as possible. The adversary does not have a particular solution in mind all answers are chosen to reveal as little information as possible while being consistent with earlier answers. The analysis proceeds by Determining what information the algorithm needs to solve the problem. Determining how long it will take to get that information from the adversary.

6 Adversary for Largest/Smallest In order to determine both the largest and the smallest we assign states to keys: X Key has not been involved in comparison. (0 unit) L Key has lost at least one comparison. (1 unit) W Key has won at least one comparison. (1 unit) WL Key has won and lost a comparison. (2 units) We cannot determine the largest and smallest keys until All keys except one have lost a comparison: n 1 units. All keys except one have won a comparison: n 1 units. We need to learn 2n 2 things overall. We design an adversary that reveals as little as possible.

7 Adversary for Largest/Smallest Before Winner After Information X X 1 W L 2 X L 1 W L 1 X W 2 L W 1 X WL 1 W WL 1 L L 1 WL W 1 L W 2 L W 0 L WL 2 L WL 0 W W 1 W WL 1 W WL 1 W WL 0 WL WL be consistent W WL 0

8 Adversary for Largest/Smallest We need to get 2n 2 items of information. How many comparisons are necessary? The most information per comparison (2) occurs when neither item has one or lost. This can happen at most n/2 times for a total of n items of information. Any other comparison results in at most 1 unit of information. We need at least 2n 2 n = n 2 of these. Total number of comparisons is n 2 + n/2 = 3n/2 2. This lower bound matches the worst case of our algorithm.

9 Second Largest Item One possibility is to find the largest, remove it, then find the largest remaining: (n 1) + (n 2) = 2n 3 comparisons. A better alternative is the tournament approach. Hold a tournament to determine the largest key. Since the second largest key must be defeated by the largest key at some point, We look for the largest key among those keys beaten by the largest key.

10 Second Largest Analysis We will just do the analysis for powers of 2. First, how many comparisons in the tournament: n 2 n 2 n lg n = n 2 2 lg n i=1 1 2 i = n 1 Next, how many keys did the max key defeat? lg n Therefore searching for the max among the defeated keys will take lg n 1 comparisons. Total comparisons: (n 1) + (lg n 1) = n + lg n 2. If n is not a power of 2: n lg n 2 Is this optimal?

11 Second Largest Adversary For any algorithm, the largest key is involved in some number of comparisons m. Discounting the largest key, there are n 1 keys. It takes at least n 2 comparisons to find the second largest among those keys. The m comparisons with the largest key cannot count toward that n 2. Therefore the total number of comparisons required is at least m + n 2. We will construct an adversary that forces m to be at least lg n.

12 Second Largest Adversary Adversary builds a tree, and uses it to guide answers. Initially all nodes are roots (throughout a root node will represent a key that hasn't lost a comparison.) Two roots compared: If both trees are the same size, answer is arbitrary, smaller is made child of larger. If one tree is larger, the root of the smaller tree loses, and is made a child of the root of the larger tree. A root and a non root are compared. The non root is declared smaller, and trees are not changed. Two non roots are compared. Answer is consistent with previous answers, and trees are not changed.

13 Second Largest Adversary When any correct algorithm terminates, there can be at most one root. Otherwise there are two keys that never lost a comparison. The rules on the previous slide are designed so that the tree with the largest key as a root grows as slowly as possible.

14 Second Largest Adversary The question is: how many comparisons must a key have been involved in to end up at the root of the final tree? The size of the tree rooted at the largest tree at most doubles after each comparison, after the kth comparison: size k 2 size k 1 Initial size of the tree rooted at the largest tree is 1. size 0 =1 Homogeneous linear recurrence. Solution is: size k 2 k. n=size m 2 m lg n m lg n m

15 Second Largest Adversary Final result is a n lg n 2 lower bound. Recall that our tournament algorithm required comparisons. n lg n 2

16 Kth Smallest Entry We can solve this problem with order n comparisons (average case) with a small modification to quicksort. Recall, that after each call to partition, the pivot item is in its final sorted position. If the pivot item is at the kth position, then we have the kth smallest item. Here is the algorithm: Partition the data, if pivot point = k, terminate. if pivot point > k, partition the array up to pivot position. if pivot point < k, partition the array after the pivot position. Repeat until pivot position = k.

17 Selection Analysis Worst case comparisons? Average case comparisons?

18 Selection Analysis Worst case comparisons? Same as quicksort: n 2. Average case comparisons? In order to do an average case analysis we sum up the cost associated with every possible input, divide by the number of possible inputs. Assume that every k is equally possible, and every pivot point is equally possible after a partition.

19 Selection Analysis Average Case Input size of recursive calls can be anything from 0 to n 1. Size is 0 if k = p (pivot point) after first partition. There are n ways for that to happen. Size is 1 if k = 1 and p = 2, or if k=n and p = n 1... two possible ways. Four ways for size to be (n 1) ways for size to be n 1.

20 Selection Analysis Average Case This leads to an ugly recurrence: A n = na 0 2 A 1 4 A 2 2 n 1 A n 1 n 1 n n 1 The book informs us that this works out to: A n 3 n

21 Selection in Worst Case Linear Time In our previous algorithm we wanted the pivot item to be near the median. This causes the size of the array to be roughly halved on each recursive call. No obvious way to select a pivot item near the median, without knowing what the median is. Determining the median requires solving the selection problem. Doh. Amazingly, there is a way forward...

22 Linear Time Selection Alter our partition function as follows: First break the array into n/5 groups. Compute the median of each of those groups. This can be done with six comparisons per group. Alternately, just use insertion sort on each group. Now, recursively call the selection algorithm to determine the median of our n/5 medians. It turns out that this median of medians will be reasonably close to the median. Use the result as the pivot point and partition as usual. The resulting algorithm is worst case linear time.

23 Linear Selection Analysis First, how close is the median of medians to the median? Let's draw a picture... So, roughly speaking, there are at most 3(n/10) items larger than the median of medians, and at most that many smaller. So the worst case size of the input to the recursive call is about 7n/10. Book comes up with 7n/10 3/2.

24 Linear Selection Analysis We can write the following recurrence to describe the worst case number of comparisons: W n = W 7 n W n 5 6 n 5 n Worst case cost of the next recursive call cost of finding the median of n/5 medians cost of finding the n/5 medians cost of partition

25 Linear Selection Analysis We don't really have any tools for dealing with this recurrence: Revert to guess and check. We guess that it is W(n) is linear, i.e. W(n) <= cn. Therefore: W n = W 7 n 10 3 W n 11 n c 7 n 10 3 c n n 5 Solving this, we get 22 <= c. c n We then use induction to prove that W(n) <= 22n.

26 Linear Selection Analysis There are selection algorithms that are closer to 3n in the worst case. We could construct an adversary argument to provide a lower bound for this problem. The highest lower bound determined so far is a bit more than 2n. Computer Algorithms, Baase and Van Gelder, 2000.